whats the official answer?

I get 10 hours? If this is right then I'll explain my approach!

That Sounds about right.

[rate pump1 + rate of pump2 ] = 1/T;
1/x + 1/1.5x = 1/4;
x = 20/3;
so the faster pump can do it in 1.5 * 20/3 = 10 hrs;

vijaynaik wrote:[rate pump1 + rate of pump2 ] = 1/T;
1/x + 1/1.5x = 1/4;
x = 20/3;
so the faster pump can do it in 1.5 * 20/3 = 10 hrs;

What do you use for the common denominator?

RickH wrote:Please help me with the following problem:

Two Water pumps working simultaneously at their respective rates, took exactly 4 hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

An easy approach would be to plug in a value for the rate of each pump:

Slow pump = 2/hour
Fast pump = (1.5)*2 = 3/hour
Combined rate = 2+3 = 5/hour
Work = rate*time = 5*4 = 20 (the value of the pool)
Time for faster pump = work/rate = 20/3.

GMATGuruNY wrote:

RickH wrote:Please help me with the following problem:

Two Water pumps working simultaneously at their respective rates, took exactly 4 hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

An easy approach would be to plug in a value for the rate of each pump:

Slow pump = 2/hour
Fast pump = (1.5)*2 = 3/hour
Combined rate = 2+3 = 5/hour
Work = rate*time = 5*4 = 20 (the value of the pool)
Time for faster pump = work/rate = 20/3.

I have not though of solving the problem in this fashion. Does this work for every problem of this kind?

i should have checked my answer before i confirmed it

Thanks 'GMATGuruNY' for the easier approach.

Gurpinder wrote:whats the official answer?

I get 10 hours? If this is right then I'll explain my approach!

A small suggestion here : Plz share ur work so that certainly improves ur confi and makes others to help u out if u make any mistakes...

It wont take minutes to post ur explanation which at end of day gonna help u and the community...

Plz excuse me if i am way off limits...
Coming to the problem,

Let A take X hours to pump

B is the faster pump and it is 1.5 time efficient than A
so B shall take 2/3 of X hours

So we shall calculate how much of water would have been pumped in 1 hour

1/X + 1/(2/3)X = 1/4

1/X + (3/2X) = 1/4
Solving for x ; X = 10

we know that B is faster and it takes (2/3) * 10 = 6.67 hrs alone

gmatmachoman wrote:

Gurpinder wrote:whats the official answer?

I get 10 hours? If this is right then I'll explain my approach!

A small suggestion here : Plz share ur work so that certainly improves ur confi and makes others to help u out if u make any mistakes...

It wont take minutes to post ur explanation which at end of day gonna help u and the community...

Plz excuse me if i am way off limits...

NP. Your excused.

I used the table method. I am using a library computer so I don't really have access to a program where I can make that chart otherwise I'd post it here.

@GMATGuruNY, is this approach correct?

4 * [x + 1.5x] = 1;
10x =1; x = 1/10; fatster machine rate 3/2 * 1/10 = 3/20; So it can do it in 20/3 hours.

Thanks

Vijaynaik,

That is the same approach i used... However i really like Rick H approach... but damn i never spot when to use the smart numbers on these problems... Oh well just need to keep practicing!!

HPengineer, that's why i want to make sure my original approach is correct. It will be useful when we don't know or forget to plugin numbers even after lot of practice.

By looking at the solution given by GMATGuruNY, it appears that my approach is correct[ W = Rate * Time].
In my 1st example I inverted the rates x and 1.5x where as i shouldn't have done that becuase x is already a RATE and not in hours.

Thanks

RickH wrote:Please help me with the following problem:

Two Water pumps working simultaneously at their respective rates, took exactly 4 hours to fill a certain swimming pool. If the constant rate of one pump was 1.5 times the constant rate of the other, how many hours would it have taken the faster pump to fill the pool if it had worked alone at its constant rate?

Let 2 pumps are A and B .

A takes X h to fill.
As B is 1/2 times faster then A hence it will take 1/3h less time to fill .
Hence B will take 2X/3 h .
now As given .... 1/x + 3/2x = 1/4 .
X=10 .

hence B (faster one ) will take 2*10/3 h = 20/3 h to fill.


Sanjeev